# PRACTICE FINAL SOLUTIONS ```PRACTICE FINAL SOLUTIONS
1. Tangent line x = 6 + 2t, y = −2 − t, z = 9 + 6t, perpendicular plane 2(x − 6) − (y +
2) + 6(z − 9) = 0
2. See Graphs at end. The surface is a hyperboloid of one sheet. The graph of the surface
is pretty much exactly Figure 8 on pg. 605 of your book.
3. Approach along y-axis = −1, Approach along x-axis = 1
√
29 &lt; −4, 8 &gt;, rate of increase in other direction =
4. Fastest
increase
direction
=
1/
√
32/ 29
5. −1(x + 2) + 2(y − 1) − 2/3(z + 3) = 0
6. Local
max
achieved along the line y = x, global min of −1 achieved
√ (and global)
√
√ of 1 √
at ( 2/2, − 2/2), (− 2/2, 2/2)
7. 44
8. 3π/2
9. 1/3(e3 − 1)
10. 16/5π
11. Graph included at end. Note that the vectors corresponding to points x = 0 do not
exist because of division by 0.
12. div = 1, curl = 0
13. 320
14. −59/105
15. Potential function f (x, y) = x2 cos y − y sin x + C, integral = π 2
16. Potential function f (x, y) = 3x2 + 5xy + 2y 2 + C, integral = 0
17. e − 1
18. The first calculation represents flux of the vector field through the unit circle, and
is 2π. The second calculation represents circulation (or rotation) of the vector field
around the unit circle, and is 0. The fact that this is 0 is a result of the fact that the
vector field is orthogonal to the unit circle at each point, so there is no tendency to
rotate.
√
√
19. Surface area = π/6(17 17 − 1), Mass = π/60(391 17 + 1)
1
20. −4/3π
21. π/2
22. 0
2
```